The domain of definition of the function `y=3e^(sqrt(x^(2)-1))log(x-1)` is

Options

A. `(1,oo)`
B. `[1,oo)`
C. `R-{1}`
D. `(-oo,-1)uu(1,oo)`

Answer

To find the domain of the function \( y = 3e^{\sqrt{x^2 - 1}} \log(x - 1) \), we need to consider the constraints imposed by both the exponential and logarithmic components of the function.

### Step-by-Step Solution:

1. **Identify the logarithmic condition:**
The logarithm \( \log(x - 1) \) is defined when its argument is positive. Therefore, we need:
\[ x - 1 > 0 \implies x > 1 \]

2. **Identify the square root condition:**
The expression \( \sqrt{x^2 - 1} \) is defined when the argument is non-negative. Thus, we need:
\[ x^2 - 1 \geq 0 \]
This can be factored as:
\[ (x - 1)(x + 1) \geq 0 \]
To solve this inequality, we find the critical points, which are \( x = -1 \) and \( x = 1 \). We can test the intervals determined by these points:
- For \( x < -1 \): both factors are negative, so the product is positive.
- For \( -1 < x < 1 \): one factor is negative and the other is positive, so the product is negative.
- For \( x > 1 \): both factors are positive, so the product is positive.

Thus, the solution to the inequality \( (x - 1)(x + 1) \geq 0 \) is:
\[ x \leq -1 \quad \text{or} \quad x \geq 1 \]

3. **Combine the conditions:**
We now have two conditions:
- From the logarithmic condition: \( x > 1 \)
- From the square root condition: \( x \leq -1 \) or \( x \geq 1 \)

The intersection of these conditions is:
\[ x > 1 \]

4. **Final Domain:**
Therefore, the domain of the function \( y = 3e^{\sqrt{x^2 - 1}} \log(x - 1) \) is:
\[ \boxed{(1, \infty)} \]